Edexcel IAL Maths
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What a parameter is; plotting a curve from a table of t-values and reading its direction
Eliminating the parameter algebraically to get a Cartesian equation
Eliminating the parameter with trigonometric identities: sin² + cos² = 1, double-angle formulae, sec² = 1 + tan², csc² = 1 + cot²
The domain and range of the Cartesian equation when the parameter is restricted
dy/dx = dy/dt ÷dx/dt; horizontal and vertical tangents
Equations of tangents and normals to a parametric curve
The area under a parametric curve, ∫ydx/dt dt, with the limits changed to values of t
1. A parametric curve gives both coordinates in terms of a third variable, the parameter (usually t or θ), as x = f(t), y = g(t), so each value of t gives one point and the curve has a direction.
2. To get a Cartesian equation, eliminate t: make t the subject of one equation and substitute, or (with trigonometric functions) use an identity that links the two expressions.
3. Domain and range: the Cartesian equation may describe more points than the curve. The x-values and y-values the curve actually reaches come from the allowed values of t.
4. Gradient: dy/dx = dy/dt ÷dx/dt. Tangent horizontal where dy/dt = 0; vertical where dx/dt = 0.
5. Area: ∫y dx = ∫ydx/dt dt, with the x-limits replaced by the matching t-limits.
6. None of these results is in the formulae list. Learn them.
Instead of linking y directly to x, both coordinates are given as functions of a third variable, the parameter: x = f(t), y = g(t). Each value of t gives exactly one point (f(t), g(t)). As t increases the point moves along the curve, so a parametric curve has a direction.
A parametric curve can do things that y = f(x) cannot: turn back on itself, have vertical tangents, or cross itself. To sketch a curve, make a table of t, x and y, plot the points in order of t and join them smoothly.
| t | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| x = t² | 4 | 1 | 0 | 1 | 4 |
| y = t³ - 3t | −2 | 2 | 0 | −2 | 2 |
The most familiar parametric curve is the path of a projectile. Launched with horizontal speed 12 m s⁻¹ and vertical speed 9 m s⁻¹, its position after t seconds is x = 12t, y = 9t - 4.9t². Time is the parameter: it tells you when the projectile is at each point, which a Cartesian equation cannot. Eliminating t = x/12 gives the path y = ¾ x - (49/1440)x².
Common error
Joining the points in order of x instead of in order of t. In the table above t = -1 and t = 1 both give x = 1, but they are different points, (1, 2) and (1, -2), on different parts of the curve.
A Cartesian equation links x and y only. To find it, you eliminate the parameter. When one equation is easy to rearrange (often a linear one such as x = 2t - 1), make t the subject and substitute into the other equation, then simplify into the form asked for. A fraction inside a fraction is cleared by multiplying top and bottom by the same thing.
Sometimes it is quicker to combine the equations directly. For example, if x = t² and y = t³ - 3t = t(t² - 3), then squaring gives y² = t²(t² - 3)² = x(x - 3)².
WORKED EXAMPLE · QUESTION
A curve C has parametric equations x = 2t - 1, y = t/(t+2), t ≠-2. (a) Show that a Cartesian equation of C is y = (x+1)/(x+5). (b) Find the coordinates of the point where C crosses the y-axis.
SOLUTION
(a) Make t the subject of the simpler equation: x = 2t - 1 ⇒t = (x+1)/2. Substitute, then multiply top and bottom by 2: y = ((x+1)/2)/((x+1)/2 + 2) = (x+1)/(x+1+4). y = (x+1)/(x+5) as required.
(b) On the y-axis x = 0, so t = ½ and y = (1/2)/(1/2 + 2) = ⅕. (0, ⅕) Check with the Cartesian equation: (0+1)/(0+5) = ⅕.
You cannot sensibly make t the subject of x = 3sin t (you would get inverse functions). Instead, find an identity that connects the two trigonometric expressions. First isolate the trigonometric function in each equation, for example x = 2 + 3cos θ⇒cos θ= (x-2)/3; then substitute into the identity. Pick the identity by looking at what appears:
| The equations contain… | Use the identity | Typical result |
|---|---|---|
| sin t and cos t | sin² t + cos² t = 1 | a circle or an ellipse |
| sin t (or cos t) and cos 2t | cos 2t = 1 - 2sin² t = 2cos² t - 1 | a parabola |
| sin t or cos t and sin 2t | sin 2t = 2sin tcos t, then sin² + cos² = 1 | often y² = … |
| tan t and sec t | sec² t = 1 + tan² t | a hyperbola-like curve |
| cot t and csc t | csc² t = 1 + cot² t | a hyperbola-like curve |
WORKED EXAMPLE · QUESTION
(a) Find a Cartesian equation of the curve x = 2 + 3cos θ, y = 3sin θ- 1, 0 ≤θ< 2π, and describe the curve. (b) A curve has parametric equations x = 3sin t, y = 4cos 2t, -π/2 ≤t ≤π/2. Show that its Cartesian equation is y = 4 - (8/9) x².
SOLUTION
(a) Isolate: cos θ= (x-2)/3, sin θ= (y+1)/3. Use sin²θ+ cos²θ= 1 and multiply by 9. (x-2)² + (y+1)² = 9: a circle with centre (2, -1) and radius 3
(b) cos 2t is linked to sin t by a double-angle formula: y = 4(1 - 2sin² t) = 4 - 8sin² t. Substitute sin t = x/3: y = 4 - 8 ×x²/9. y = 4 - (8/9) x² as required. Choose cos 2t = 1 - 2sin² t because x involves sin t; the version with cos² t would need a second step.
Common error
Squaring and adding before isolating the trigonometric functions. From x = 2 + 3cos θ and y = 3sin θ- 1, writing x² + y² = 9 is wrong: the constants 2 and −1 must be moved to the other side first.
The Cartesian equation can describe more than the parametric curve. In worked example 2(b), y = 4 - (8/9) x² is a whole parabola, but the curve is only the part with -3 ≤x ≤3. So a complete answer gives the Cartesian equation together with its domain, the set of x-values the curve reaches. The range is the set of y-values it reaches.
To find them, work out the values that x = f(t) and y = g(t) take over the allowed values of t. Use what you know about each function: -1 ≤sin t ≤1, eᵗ > 0, t² ≥0, sec t ≥1 for 0 ≤t < π/2, and so on. The restriction on t can also decide a sign: if y² = … and y is known to be positive, take the positive square root.
WORKED EXAMPLE · QUESTION
A curve C has parametric equations x = tan t + 1, y = 3sec t, 0 ≤t < π/2. (a) Find a Cartesian equation of C in the form y = √(f(x)), where f(x) is a quadratic. (b) State the set of values of x (the domain) and of y (the range) on C.
SOLUTION
(a) Isolate: tan t = x - 1, sec t = y/3. Use sec² t = 1 + tan² t: y²/9 = 1 + (x-1)² ⇒y² = 9x² - 18x + 18. For 0 ≤t < π/2, sec t > 0, so y > 0. y = √(9x² - 18x + 18)
(b) As t runs from 0 towards π/2, tan t increases from 0 without limit and sec t increases from 1 without limit. Domain x ≥1; range y ≥3
Exam tip 1 — the domain comes from x = f(t)
When a question says "state the domain of f" after you have found y = f(x), the answer comes from the parametric equation for x, not from the formula for f. A formula like √(9x² - 18x + 18) is defined for every x, but the curve only has x ≥1.
Gradient of a parametric curve
dy/dx = dy/dt ÷dx/dt (provided dx/dt ≠0)
Differentiate x and y separately with respect to t, then divide; the answer is in terms of t. It follows from the chain rule: dy/dx = dy/dt ×dt/dx, and dt/dx = 1 ÷dx/dt. Not in the formulae list, so learn it.
To find the gradient at a point you need its value of t. If you are given the coordinates, find the t that gives both of them.
Horizontal tangent (gradient 0): dy/dt = 0 with dx/dt ≠0.
Vertical tangent (gradient undefined): dx/dt = 0 with dy/dt ≠0.
Simplify the quotient, especially with trigonometric functions: -2cos θ/4sin θ = -½cot θ.
WORKED EXAMPLE · QUESTION
The curve C has parametric equations x = t², y = t³ - 3t. (a) Find dy/dx in terms of t. (b) Find the gradient of C at the point (4, 2). (c) Find the coordinates of the points where the tangent to C is horizontal, and of the point where it is vertical.
SOLUTION
(a) dx/dt = 2t, dy/dt = 3t² - 3. dy/dx = (3t² - 3)/2t
(b) Find t at (4, 2) using both coordinates: t² = 4 gives t = ±2; y = 2 only for t = 2 (t = -2 gives y = -2). Gradient = (3×4 - 3)/(2×2). (9/4)
(c) Horizontal where dy/dt = 0: 3t² - 3 = 0 ⇒t = ±1, and dx/dt = ±2 ≠0 there. (1, -2) and (1, 2) Vertical where dx/dt = 0: 2t = 0 ⇒t = 0, with dy/dt = -3 ≠0. (0, 0)
At the crossing point (3, 0) the two branches have gradients (3×3 - 3)/(±2√3) = ±√3: two different tangents through one point, which no curve y = f(x) can have.
The method is the one from P1; only the gradient is found differently.
Find the point: substitute the value of t into both parametric equations.
Find the gradient m: substitute the same t into dy/dx. For the normal use -1/m.
Use y - y₁ = m(x - x₁), with x and y, never t, in the final equation.
WORKED EXAMPLE · QUESTION
The ellipse E has parametric equations x = 4cos θ, y = 2sin θ, 0 ≤θ< 2π. The point P on E has parameter θ= π/4. (a) Show that dy/dx = -½cot θ. (b) Find an equation of the tangent to E at P, in the form x + ay = b, where a and b are exact constants. (c) Find an equation of the normal to E at P. (d) The tangent meets the x-axis at A and the y-axis at B. Find the area of triangle OAB.
SOLUTION
(a) dx/dθ = -4sin θ, dy/dθ = 2cos θ, so dy/dx = 2cos θ/(-4sin θ). -½cot θ as required.
(b) At θ= π/4: P = (4×√2/2, 2×√2/2) = (2√2, √2) and m = -½cot π/4 = -½. Then y - √2 = -½(x - 2√2) ⇒2y - 2√2 = -x + 2√2. x + 2y = 4√2
(c) Normal gradient -1 ÷(-½) = 2: y - √2 = 2(x - 2√2). y = 2x - 3√2
(d) Tangent intercepts: y = 0 gives x = 4√2, so A(4√2, 0); x = 0 gives y = 2√2, so B(0, 2√2). Area = ½ ×4√2 ×2√2 = ½ ×16. 8
Common error
Leaving θ (or t) in the equation of a tangent. The gradient -½cot θ is a formula; the tangent at P uses the number -½. A final answer such as y - √2 = -½cot θ (x - 2√2) is not a straight line.
Area under a parametric curve
Area = ∫ₓ₌ₐˣ⁼ᵇ y dx = ∫ₜ₌ₚ^(t=q) y dx/dt dt
p and q are the values of t at which x = a and x = b; y and dx/dt are both written in terms of t. It is integration by substitution: dx is replaced by dx/dt dt, and the limits change with it. Not in the formulae list, so learn it.
The curve must be above the x-axis for the result to be the area. Keep the limits in the order "t at the left end, x = a" to "t at the right end, x = b", even if the first is the larger number. The method:
find the t-values at the ends of the region (from the x-values given, or from where the curve meets an axis);
differentiate x with respect to t;
multiply y by dx/dt and simplify (this is often the step a "show that" part is testing);
integrate with respect to t between the t-limits.
If x decreases as t increases, dx/dt is negative and the left-hand end has the larger t. Writing the limits in the correct order makes the answer positive; you can then swap them and change the sign of the integrand, as in worked example 7.
WORKED EXAMPLE · QUESTION
The curve C has parametric equations x = t² + 2t, y = 3 - t, 0 ≤t ≤3. Find the area of the region bounded by C and the coordinate axes.
SOLUTION
Ends of the region. t = 0 gives (0, 3), on the y-axis; t = 3 gives (15, 0), on the x-axis. x increases from 0 to 15 as t goes from 0 to 3.
Set up. dx/dt = 2t + 2, so Area = ∫₀³ (3 - t)(2t + 2) dt = ∫₀³ (6 + 4t - 2t²) dt.
Integrate. [6t + 2t² - ⅔ t³]₀³ = 18 + 18 - 18 - 0.
18
WORKED EXAMPLE · QUESTION
The curve C has parametric equations x = 4cos t, y = sin 2t, 0 ≤t ≤π/2. The region R is bounded by C, the x-axis and the line x = 2. (a) Show that the area of R is given by ∫_(π/3)^(π/2) 8sin² tcos t dt. (b) Hence find the exact area of R.
SOLUTION
(a) R runs from x = 0 to x = 2. x = 0: cos t = 0, t = π/2; x = 2: cos t = ½, t = π/3. With the limits in x-order and sin 2t = 2sin tcos t: Area = ∫_(π/2)^(π/3) sin 2t ×(-4sin t) dt = ∫_(π/2)^(π/3) -8sin² tcos t dt.
Swap the limits and change the sign: Area = ∫_(π/3)^(π/2) 8sin² tcos t dt as required.
(b) Integrate by recognition: the derivative of sin³ t is 3sin² tcos t. [(8/3)sin³ t]_(π/3)^(π/2) = (8/3)(1 - 3√3/8), using sin³π/3 = 3√3/8. (8/3) - √3 (≈ 0.935)
Common error
Putting the x-values 0 and 2 in as the limits of a t-integral. Once the integrand is in terms of t, the limits must be values of t (here π/2 and π/3). Mixing them spoils an otherwise correct integral.
Longer questions on this topic usually combine a Cartesian equation (with its domain), a tangent or normal, and an area. The last example does all three.
WORKED EXAMPLE · QUESTION
The curve C has parametric equations x = tan t, y = sin 2t, 0 ≤t ≤π/4.
(a) Show that a Cartesian equation of C is y = 2x/(1+x²), and state the domain of x on C.
(b) Find dy/dx in terms of t.
(c) The point P on C has parameter t = π/6. Find an equation of the tangent to C at P, in the form y = mx + c, where m and c are exact constants.
(d) Find the exact area of the region bounded by C, the x-axis and the line x = 1.
SOLUTION
(a) Write sin 2t in terms of tan t: sin 2t = 2sin tcos t = 2sin t/cos tcos² t = 2tan t/(sec² t) = 2tan t/(1 + tan² t). So y = 2x/(1+x²); tan t increases from 0 to 1 on 0 ≤t ≤π/4, so the domain is 0 ≤x ≤1.
(b) dx/dt = sec² t, dy/dt = 2cos 2t, so dy/dx = 2cos 2t/(sec² t). 2cos 2tcos² t
(c) P = (tan π/6, sin π/3) = (√3/3, √3/2) and m = 2 ×½ ×(√3/2)² = ¾. Then y - √3/2 = ¾(x - √3/3) ⇒y = ¾ x - √3/4 + √3/2. y = ¾ x + √3/4
(d) x = 0 at t = 0 and x = 1 at t = π/4. Area = ∫₀^(π/4) sin 2tsec² t dt = ∫₀^(π/4) 2sin tcos t/(cos² t) dt = ∫₀^(π/4) 2tan t dt = [2ln sec t]₀^(π/4) = 2ln √2 - 2ln 1. ln 2 As a check, ∫₀¹ 2x/(1+x²) dx = [ln(1+x²)]₀¹ = ln 2 as well.
Exam tip 2 — label your limits
Write the change of limits as a separate line: "x = 0 ⇒t = π/2, x = 2 ⇒t = π/3". It shows the step clearly on its own, and it stops you putting x-values into a t-integral.
Exam tip 3 — in "show that" area parts, show the product
If the target is an integral such as ∫8sin² tcos t dt, the working must show y ×dx/dt written out, any identity used (sin 2t = 2sin tcos t), and how the limits and sign were sorted out. Jumping straight to the printed answer proves nothing.
Exam tip 4 — one value of t must fit both coordinates
When you are given a point rather than a parameter, solve one equation for t and check each possible value in the other. With x = t², both t = 2 and t = -2 give x = 4, but only one of them gives the right y.
Exam tip 5 — simplify dy/dx, but leave it in terms of t
"Find dy/dx in terms of t" is complete once the quotient is simplified (-½cot θ, 2cos 2tcos² t). Do not try to convert it into x; that wastes time and invites errors.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| dy/dx = dx/dt ÷dy/dt. | That is dx/dy, the reciprocal of the gradient. | dy/dx = dy/dt ÷dx/dt: the variable you want on top goes on top. |
| The area under a parametric curve is ∫y dt. | The area is ∫y dx; changing the variable to t changes dx as well. | Area = ∫ydx/dt dt. |
| The limits of the t-integral are the x-values at the ends of the region. | After substitution everything is in terms of t, including the limits. | Convert each x-limit to its value of t first. |
| The Cartesian equation describes exactly the same points as the parametric equations. | The parameter's range can restrict the curve (-1 ≤sin t ≤1 keeps |x| ≤3 when x = 3sin t). | Give the Cartesian equation with the domain the parametric curve actually covers. |
| A vertical tangent occurs where dy/dt = 0. | dy/dt = 0 makes the gradient zero, so the tangent is horizontal. | Vertical tangent: dx/dt = 0 (with dy/dt ≠0). |
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| If x = 2 + 3cos θ and y = 3sin θ, then x² + y² = 9. | The identity applies to cos θ and sin θ, not to x and y when constants have been added. | Isolate first: cos θ= (x-2)/3, so (x-2)² + y² = 9. |
P3 Trigonometry and differentiation
Eliminating a parameter relies on the P3 identities (double-angle formulae, sec² = 1 + tan², csc² = 1 + cot²), and dy/dx = dy/dt ÷dx/dt is the chain rule. Tangents and normals use the straight-line work of P1; circles from P2 reappear as parametric curves.
P4.4 Implicit differentiation and P4.6 Integration
Implicit differentiation is the other way to differentiate a curve that is not written as y = f(x). The area formula is integration by substitution, and volumes of revolution use the same idea: π∫y²dx/dt dt.
Mechanics and IAL Physics
A projectile's position is a parametric curve with time as the parameter: x = (ucos α)t, y = (usin α)t - ½ gt². Eliminating t gives the equation of the path, and dy/dx gives the direction of motion. Position vectors such as r = f(t)i + g(t)j are the same idea, and projectile motion in IAL Physics Unit 1 treats horizontal and vertical motion separately in the same way.
| Formula | Use | In the formulae list? |
|---|---|---|
| dy/dx = dy/dt ÷dx/dt | gradient of a parametric curve | No — learn it |
| Area = ∫ydx/dt dt, between t-limits | area between a parametric curve and the x-axis | No — learn it |
| sin² t + cos² t = 1 | eliminating t from sin and cos (circles, ellipses) | No — learn it |
| sin 2t = 2sin tcos t; cos 2t = 1 - 2sin² t = 2cos² t - 1 | eliminating t when a double angle appears | No — learn it |
| sec² t = 1 + tan² t; csc² t = 1 + cot² t | eliminating t from tan and sec, or cot and cosec | No — derive it from sin² + cos² = 1 |
| y - y₁ = m(x - x₁); mₙₒᵣₘₐₗ = -1/m | equations of tangents and normals | No — learn it |
| ∫tan t dt = ln|sec t| + c | a standard integral that is useful in parametric areas | Yes |
1. Find a Cartesian equation of the curve x = 3t - 1, y = 2/t, t ≠0.
2. Show that the curve x = 1 + 2cos t, y = 2sin t - 3 is a circle, and state its centre and radius.
3. A curve has x = eᵗ, y = e²ᵗ + 1, t ∈ℝ. Find its Cartesian equation, its domain and its range.
4. For x = t³, y = t² + t, find the gradient of the curve at the point where t = 1.
5. A student finds the area under a parametric curve and gets a negative answer, although the curve is above the x-axis. What has probably gone wrong?
6. For the ellipse x = 4cos θ, y = 2sin θ, find the points where the tangent is horizontal.
1. t = (x+1)/3, so y = 2 ÷(x+1)/3 = 6/(x+1), x ≠-1.
2. cos t = (x-1)/2 and sin t = (y+3)/2. Then sin² + cos² = 1 gives (x-1)² + (y+3)² = 4: centre (1, -3), radius 2.
3. e²ᵗ = (eᵗ)² = x², so y = x² + 1. Since eᵗ > 0, the domain is x > 0, and then the range is y > 1. It is only the right half of the parabola, without its vertex.
4. dy/dx = (2t+1)/3t² = (3/3) = 1 at t = 1 (the point (1, 2)).
5. x decreases as t increases, so dx/dt < 0, and the student has written the smaller t as the lower limit. The lower limit should be the value of t at the left-hand end of the region (the smaller x). With the limits in that order, the answer comes out positive.
6. dy/dθ = 2cos θ= 0 when θ= π/2 or 3π/2, and dx/dθ = -4sin θ≠0 there. The points are (0, 2) and (0, -2), the top and bottom of the ellipse.
Check understanding
The answer and each step appear one click at a time.
A curve C has parametric equations x = 3t + 2, y = t² − 1, t ∈ ℝ.
(a) Find the coordinates of the point on C where t = 2. [1]
(b) Find a Cartesian equation of C in the form y = ax² + bx + c, where a, b and c are constants. [3]
(c) Find the x-coordinates of the points where C meets the x-axis. [2]
Mark scheme click to reveal each step
(8, 3)
t = (x − 2)/3
substitutes: y = ((x − 2)/3)² − 1
y = (1/9)x² − (4/9)x − 5/9
y = 0 ⇒ t² = 1 ⇒ t = ±1, or solves x² − 4x − 5 = 0
x = 5 and x = −1
A curve C has parametric equations x = 5 cos θ − 2, y = 5 sin θ + 3, 0 ≤ θ ≤ π.
(a) Show that C is part of the circle with equation (x + 2)² + (y − 3)² = 25. [3]
(b) State the centre and the radius of this circle. [2]
(c) Find the set of values of x and the set of values of y on C. [2]
Mark scheme click to reveal each step
cos θ = (x + 2)/5 and sin θ = (y − 3)/5
uses sin² θ + cos² θ = 1
((x + 2)/5)² + ((y − 3)/5)² = 1 ⇒ (x + 2)² + (y − 3)² = 25 as required
centre (−2, 3)
radius 5
cos θ takes all values from 1 down to −1, so −7 ≤ x ≤ 3
sin θ takes all values from 0 to 1, so 3 ≤ y ≤ 8
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